I can kind of picture it in my head, the region is basically above the xy plane, and thus, only the positive root of z = sqrt(16 - (x^2 + y^2)) needs to be taken into account. I also know that both the cone and the sphere intersect the z axis at (0,0,4) and this is also and intersection point for the curves.

Part B:
Clearly identify the limits of integration using spherical coordinates and set up the iterated triple integral which gives the volume bounded by the above sketch. Do not evaluate the integral!

So the limits of integration are:
Since there is a circle in the xy plane of x^2 + y^2 = 16, we can integrate with respect to theta from 0 to 2pi.
For phi, we determined above that the upper limit was pi/2, since the region is above the xy plane, the lower limit is 0.
We also determined that row had a value of 4 which is the upper limit and the lower limit is 0.
So the iterated triple integral excluding the limits is:
integral d(theta) integral(sin(phi)d(phi) integral((row^2)*(4 - row(sin(phi)))d(row). Right?!

Part C:
Evaluate the volume given by the iterated triple integral above using pappus theory:
V(E) = 2pi*centre of mass y*A(D) = the volume of E is the area of D times the circumference o fthe circle traversed by the centroid (x,y) of D.
I am confused because Pappus theory is supposed to simplify volumes. However, in order to find the centroid, I need to find the centre of mass first and the moment of the lamina with respect to the x-axis.

Subtract from the volume of the top half of the sphere and from volume of cone.
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Hope the diagram helps. I am not familiar with any 3-d plot. I recommend you to know the quadric surfaces very well and know transformations.

Let me explain I know that was a cone. First, is a cone along the x-axis (you need to know that from quadric surfaces). But you have a minus in front. Thus, it is an upsidedown cone (remember mirror images are negatives of the function, that is what I mean by a transformation). But you have a 4 in from. Thus you are going to shift the cone 4 units up the z-axis and there you have it.

I know what you are saying and I have done this to verify the volume I get by pappus. Unfortunately, I am encouraged to use pappus only. Can you tell whether I have set up the limits and the iterated triple integral correct?

Thank you for your quick reply; I wish it was as simple as subtraction of the volumes as you suggested. However, this is good as it verifies that my depiction of the surfaces is correct.

Could you please verify my triple iterated integral and its limits for the question in my first post regarding the sphere and the cone; essentially part B of the question.

Do you know how Pappus can simplify finding the volume of of the region below the sphere and outside the cone. I know you suggested to subtract the volume of cone from the volume of the sphere; but the question strictly asks for Pappus (part C).

I managed to get the volume for the region below the sphere and above the cone by Pappus.

I subtracted the cone volume from the volume of half of the sphere.

V sphere = (4r^3pi)3
V half sphere = (2r^3pi)/3
V cone = (h*pi*r^2)/3
Both the cone and the sphere share a radius of 4, and the cone height is also 4, thus:

V solid = (2r^3pi)/3 - (h*pi*r^2)/3 = (64(2pi - pi))/3 = 64pi/3

I am confident that this is the volume of interest. However, I cannot get the same volume from the triple iterated integral and the limits I suggested in my first reply. Since I cannot get the same volume from the integral I proposed as the volume I calculated by Pappus, my proposed triple iterated integral and its limits must be wron, any hints why?